Geometric meaning of derivative of a complex valued function I know that a derivative of a real valued function at a point measures the slope of the tangent at that point. I am wondering what could be the geometric meaning of derivative of a complex valued function at a certain point?
 A: The derivative gives you the best linear approximation:
$$
f(z+h) \approx f(z) + f`(z)h 
$$
when $|h|$ is small.
The right side of that approximation is a straight line in the space $\mathbb{C} \times \mathbb{C}$.
(Whether you consider this "geometric" is up to you.)
A: The geometric meaning of having $f'(z_0)=0$ is that, if you consider the graphs of $\operatorname{Re}f$ and of $\operatorname{Im}f$, then the graph is smooth near $z_0$ and its tangent planes there are horizontal.
On the other hand, asserting that $f'(z_0)=c\neq0$ means that $f$ maps a tiny circle near $z_0$ ($t\mapsto z_0+re^{it}$, with $t\in[0,2\pi]$), seen as a point that moves once clockwise around $z_0$, starting tat $z_0+r$, into a small circle near $f(z_0)$, whose radius is the radius $r$ of the original circle times $\lvert c\rvert$ and whose initial point is about $f(z_0)+cr$. Actually, this circle is very close to $t\mapsto f(z_0)+cre^{it}$ and, in particular, if you see it as a point that moves around $f(z_0)$, it moves once and clockwise.
That's why the conjugation is differentiable nowhere: it reverses the orientation of the movement of the point around $f(z_0)$.
A: A holomorphic function $f$ is simply a usual $C^1$ function from, say, some open subset of $\mathbb{R}^2$ into $\mathbb{R}^2$. Say $f(z)=u(z)+iv(z)$ for real functions $u$ and $v$.
Then, $f'(x+iy)=\frac{\partial u}{\partial x}(x+iy)+i\frac{\partial v}{\partial x}(x+iy)=\frac{\partial v}{\partial y}(x+iy)-i\frac{\partial u}{\partial y}(x+iy)$ by the Cauchy-Riemann equations. Of course, switching this around, you get that really, $f'=(\partial_x u,-\partial_y u)$, so you can think of $f'$ as a twisted gradient, which also observes some compatability between the two functions $u$ and $v$.
